Quantitative Aptitude Test - 4 - PDF Flipbook

Quantitative Aptitude Test - 4

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GATE
EEE

Quantitative
Aptitude

Test-04Solutions


QUANTITATIVE APTITUDE
1. The ages of Rahul and Madhan differ by 16 years. Six years

ago, Madhan’s age was thrice as that of Rahul’s, find their
present ages
a) 14 years, 30 years
b) 12 years, 28 years
c) 16 years, 34 years
d) 18 years, 38 years
Answer: (a)
Solution:
Let Rahul’s age = x years
So, Madhan’s age = (x + 16) years
Also, 3(x - 6) = x + 16 – 6 or, x = 14
Rahul’s age = 14 years
And, Madhan’s age 14 + 16 = 30 years
2. The variable cost (V) of manufacturing a product varies
according to the equation V = 4q, where q is the quantity
produced. The fixed cost (F) of production of same product
reduces with q according to the equation F = 100/q. How many
units should be produced to minimize the total cost (V + F)
a) 5
b) 4
c) 7
d) 6
Answer: (a)

1


Solution:

V = 4q

F = 100
q

(V + F) = 4q + 100
q

(a): ‘5’

= 4 × 5 + 100 = 20 + 20 = 40
5

(a): ‘4’

= 4 × 4 + 100 = 16 + 25 = 41
5

(c): ‘7’

= 4 × 7 + 100 = 28 + 14.28 = 42.28
5

(c): ‘6’

= 4 × 6 + 100 = 24 + 16.66 = 40.66
5

∴ No. of units should be produced to minimize the total cost

(V + F) = ‘5’

3. The average price of 10 books is Rs.12 while the average price

of 8 of these books is Rs.11.75. Of the remaining two books, if

the price of one book is 60% more than the price of the other,

what is the price of each of these two books?

a) Rs. 5, Rs.7.50

b) Rs. 8, Rs. 12

c) Rs. 10, Rs. 16

d) Rs. 12, Rs. 14

Answer: (c)

2


Solution:
Total cost of 10 books = Rs. 120.
Total cost of 8 books = Rs. 94.
⇒ The cost of 2 books = Rs. 26.
Let the price of each book be x and y.
⇒ x + y = 26 ---------------- (1)
Given that the price of 1 book is 60% more than the other price
(160/100)y + y = 26.
y(160/100 + 1) = 26.
y = (26 × 100)/260.
y = 10.
Substituting y = 10 in (1) we get,
x + 10 = 26.
x = 16.
4. The calendar of the year 2024 can be used again in the year?
a) 2030
b) 2052
c) 2048
d) 2036
Answer: (b)
Solution:
Given year 2024 when divided by 4, leaves a remainder 0.
NOTE: When remainder is 0, 28 is added to the given year to
get the result.
So, 2024 + 28 = 2052

3


5. A and B are friends. They decide to meet between 1 PM and 2
PM on a given day. There is a condition that whoever arrives
first will not wait for the other for more than 15 minutes. The
probability that they will meet on that day is
a) 1/4
b) 1/16
c) 7/16
d) 9/16
Answer: (c)
Solution:
Two friend A and B

The probability that they will meet on that day

= 1 – 2�21 × 45 × 6450�
60

(Area of ABCD)

= �1 − �43 × 43��

= 1 – 9 = 16 − 9 = 7
16 16 16

4


6. Some men promised to do a job in 40 days. 12 of them were
absent and the remaining men did the job in 60 days. What is the
original number of men?
a) 44
b) 36
c) 40
d) 38
Answer: (b)
Solution:
Let there be m men in the beginning, who promised to do the
job in 40 days.
But 12 became absent; means the remaining (m – 12) men did
the job in 60 days.
The work is same ⇒ 40m = 60(m – 12) ⇒ 40m = 60m – 720
⇒ m = 36. Hence, answer is 36 men.

7. At what time between 5 and 6 will the hands of the clock
coincide?
a) 262/11 minutes past 5
b) 263/11 minutes past 5
c) 283/11 minutes past 5
d) 273/11 minutes past 5
Answer: (d)
Solution:
It’s better to use formula as it can save lots of time in exams.
However, we should understand the basics for sure. Please find

5


the method given below to solve the same problem in the

traditional way.

At 5 o' clock, the hands are 25-minute spaces apart

Hence minute hand needs to gain 25 more-minute spaces so that

the hands will be in opposite direction.

We know that 55 min spaces are gained by minute hand (with

respect to hour hand) in 60 min

Hence time taken for gaining 25-minute spaces by minute hand

= 60 ∙ 25 min ute = 12 ∙ 25 min ute = 300 min ute = 27131 minute
55 11 11

Hence hands of the clock will be together at 27131 minute past 5.

8. Find the largest number of four digits which is exactly divisible

by 27, 18, 12, 15

a) 8720

b) 9720

c) 10720

d) 11720

Answer: (b)

Solution:

LCM of 27-18-12-15 is 540.

After dividing 9999 by 540 we get 279 remainder.

So, answer will be 9999 – 279 = 9720

6


9. Raju has 14 currency notes in his pocket consisting of only Rs.

20 notes and Rs. 10 notes. The total money value of the notes is

Rs. 230. The number of Rs. 10 notes that Raju has is

a) 5

b) 6

c) 9

d) 10

Answer: (a)

Solution:

Total No. of currency notes = 14

Rs. 20 notes = x

Rs. 10 notes = y

x + y = 14 ………… (i)

Money value of Rs. 20 notes = 20 x

Money value of Rs. 10 notes = 10 y

The total money value of the notes = Rs. 230

20x + 10y = 230 ………. (ii)

Solve (i) and (ii)

x + y = 14 ………. (i) × 20

20x + 10y = 230 ………. (ii)

20x + 20y = 280

20x + 10y = 230

10y = 50

y = 50 = 5
10

∴ The number of Rs. 10 notes that has is 5.

7


10. A 25 cm wide path is to be made around a circular garden
having a diameter of 4 meters. Approximate area of the path is
square meters is
a) 3.34
b) 2
c) 4.5
d) 5.5
Answer: (a)
Solution:
Area of the path = Area of the outer circle – Area of the inner
circle = ∏{4/2 + 25/100}2 – ∏[4/2]2
= ∏[2.252 – 22] = ∏(0.25)(4.25)
{(a2 – b2 = (a – b) (a + b)}
= (3.14) (1/4)(17/4) = 53.38/16 = 3.34 sq m

11. The least common multiple of two natural numbers a and b, is
399. What is the minimum possible sum of the digits of the
number a (given a > b)?
a) 1
b) 3
c) 5
d) 7
Answer: (b)
Solution:
Factorize 399 = 3 × 7 × 19
The possible pairs are: (57, 7), (21, 19), (399, 1), (133, 3)

8


The least possible sum is given when a = 21 and b = 19
And the sum is 2 + 1 = 3
12. A, B and C enter into a partnership. They invest Rs. 40,000,
Rs. 80,000 and Rs. 1, 20,000 respectively. At the end of the first
year, B withdraws Rs. 40,000, while at the end of the second
year, C withdraws Rs. 80,000. In what ratio will the profit be
shared at the end of 3 years?
a) 2:3:5
b) 3:4:7
c) 4:5:9
d) None of these
Answer: (b)
Solution:
A: B: C = (40000 × 36): (80000 × 12 + 40000 × 24): (120000 ×
24 + 40000 × 12)

= 144:192:336 = 3:4:7
13. In an office in Singapore there are 60% female employees. 50

% of all the male employees are computer literate. If there are
total 62% employee’s computer literate out of total 1600
employees, then the no. of female employees who are computer
literate?
a) 690
b) 674
c) 672
d) 960

9


Answer: (c)
Solution:
Total employees, = 1600
Female employees, 60% of 1600. = (60 ×1600) /100 = 960.
Then male employees, = 640
50% of male are computer literate, = 320 male computer literate.
62% of total employees are computer literate, = (62 × 1600)
/100 = 992 computer literate.
Thus, Female computer literate = 992 – 320 = 672.
Alternatively:
Let 60% employees are female and 40% are male.
Then, 20% of male are computer literate and 42% are female
computer literate.
Female computer literate = (1600 × 42) /100

= 672.
14. An automobile plant contracted to buy shock absorbers from

two suppliers X and Y, X supplies 60% and Y supplies 40% of
the shock absorbers. A11 shock absorbers are subjected to a
quality test. The ones that pass the quality test are considered
reliable. Of X's shock absorbers, 960% are reliable. Of Y's
shock absorbers, 72% are reliable. The probability that a
randomly chosen shock absorber, which is found to be reliable,
is made by Y is
a) 0.288
b) 0.334

10


c) 0.667

d) 0.720

Answer: (b)

Solution:

X supplies = P(x) = 0.6

Y supplies = P(y) = 0.4

X’s shock absorbers reliable = P�RX� = 0.96

X’s shock absorbers reliable = P�RY� = 0.72

P�RY� = P(Y∩R)
P(R)

= P(Y)P�RY� = (0.4) × (0.72)
P(X)P�RX�+P(Y)P�RY� (0.6)(0.96) + (0.4)(0.72)

= 0.288 = 0.288 = 0.334
0.576 + 0.288 0.864

Alternate Method:

100

XY

60% 40%

Reliable 96% 72%

60 × 96 40 × 72
100 100

0.576 0.288

The probability that a randomly chosen shock absorber,

which is found to be reliable, is made by Y is

= 0.288 = 0.334
0.576 + 0.288

11


15. In a cricket championship, there are 21 matches. If each team
plays one match with every other team, the number of teams is
a) 7
b) 9
c) 10
d) None of these
Answer: (a)
Solution:
Let n be the number of teams.
nC2 = 21
(n (n – 1)/2) = 21
⇒ n(n – 1) = 42
∴⇒n=7

16. A political party orders an arch for the entrance to the ground
in which the annual convention is being held. The profile of the
arch follows the equation y = 2x – 0.1x2 where y is the height of
the arch in meters. The maximum possible height of the arch is
a) 8 meters
b) 10 meters
c) 12 meters
d) 14 meters
Answer: (b)
Solution:
The profile of the arch y = 2x – 0.1x2
y = height of the arch in meters

12


Assume x = 2, 4, 6, 8, 10, 12, 14, 16 ………
x y = 2x – 0.1x2
2 y = 2 × 2 – 0.1 × (2)2 = 3.6
4 y = 2 × 4 – 0.1 (4)2 = 6.4
6 y = 2 × 6 – 0.1 × (6)2 = 8.4
8 y = 2 × 8 – 0.1 × (8)2 = 9.6
10 y = 2 × 10 – 0.1 × (10)2 = 10
12 y = 2 × 12 – 0.1 × (12)2 = 9.6
14 y = 2 × 14 – 0.1 × (14)2 = 8.4
16 y = 2 × 16 – 0.1 × (16)2 = 6.4
.
.
.
.

∴ The maximum possible height of the arch is at x = 10
Alternate Method:
y = 2x – 0.1x2
y' = 2 – 0.1 × 2x ⟹ x = 10
y'' = – 2 × 0.1 ⟹ y''(10) = – 0.2 < 0 ⟹ y is max @ x = 10
17. The probability that a bowler bowled a ball from a point will
hit by the batsman is 1/4. Three such balls are bowled
simultaneously towards the batsman from that very point. What
is the probability that the batsman will hit the ball?
a) 37/64
b) 27/56

13


c) 11/13

d) 9/8

Answer: (a)

Solution:

Probability of not hitting the ball = 1 – 1/4 = 3
4

Then, the probability that the batsman will hit the ball

= 1 − �43�3 = 37
64

18. Jacob brought a scooter for a certain sum of money. He spent

10% of the cost of repairs and sold the scooter for a profit of Rs.

1100. How much did he spend on repairs if he made a profit of

20%?

a) Rs. 400

b) Rs. 440

c) Rs. 500

d) Rs. 550

Answer: (c)

Solution:

Let the C.P. be Rs. x. Then, 20% of x = 1100
20/100 × x = 1100 ⇒ x = 5500

C.P. = Rs. 5500, expenditure on repairs = 10%

Actual price = Rs. (100 × 5500)/110 = Rs. 5000

Expenditures on repairs = (5500 – 5000) = Rs. 500.

14


19. Find 2 numbers that sum to 21 and the sum of the squares is
261.
a) 14 and 7
b) 15 and 6
c) 16 and 5
d) 17 and 4
Answer : (b)
Solution:
The numbers are 15 and 6.
x + 7 = 21
⇒ x = 21 – 7
x2 + y2 = 261
(21 – 7 )2 + y2 = 261
441 – 42y + y2 + y2 = 261
2y2 – 42y + 180 = 0
y2 – 21y + 90 = 0
y1,2 = 21 + 441 – 360/2
y1,2 = 21 + 81/2
y1,2 = 21 + 9/2
y1 = 15
y2 = 6
x1 = 21 = y1 = 21 – 15 = 6
x2 = 21 – y2 = 21 – 6 = 15

15


20. In a game, A can give B 20 points, A can give C 32 points and

B can give C 15 points. How many points make the game?

a) 120 points

b) 90 points

c) 80 points

d) 100 points

Answer: (d)

Solution:

⟹ When B scores 1 point, C scores ( − 15) points


⟹ When B scores (x – 20) point, C scores ( − 20)( − 15) points


i.e. (x – 32) = ( − 20)( − 15)


⟹ x(x – 32) = ( − 20)( − 15)

⟹ x2 – 32x = x2 – 35x + 300

– 32x = – 35x + 300
⟹ 3x = 300

⟹ x = 100

i.e. 100 points make the game

21. The number of students in 3 classes is in the ratio 2:3:4. If 12

students are increased in each class this ratio changes to 8:11:14.

The total number of students in the three classes in the

beginning was

a) 162

b) 108

c) 96

16


d) 54
Answer: (a)
Solution:
Let the number of students in the classes be 2x, 3x and 4x
respectively;
Total students = 2x + 3x + 4x = 9x.
According to the question,
(2x +12)/(3x + 12) = 8/11
24x + 96 = 22x + 132
Or, 2x = 132 – 96
Or, x = 36/2 = 18
Hence, Original number of students,
9x = 9 × 18 = 162.
22. X and Y are two positive real numbers such that 2X + Y ≤ 6
and X + 2Y ≤ 8. For which of the following values of (X, Y) the
function f(X, Y) = 3X + 6Y will give maximum value?
a) (4/3, 10/3)
b) (8/3, 20/3)
c) (8/3, 10/3)
d) (4/3, 20/3)
Answer: (a)
Solution:
2x + y ≤ 6 ………. (i)
x + 2y ≤ 8 ………. (ii)
equation (i) × 2

17


4x + 2y ≤ 12 ………. (i)

x + 2y ≤ 8 ………. (ii)

––

3x ≤ 4

∴ x ≤ 34, substitute in equation no (ii)

4 + 2y ≤ 8 ⟹ 2y ≤ 8 – 4 = 24 − 4 = 20
3 3 3 3

∴ y ≤ 10 s
3

⟹ In options (b) and (c) x value is 8/3, it is > 4/3 So, option (b)

and (c) are incorrect.

⟹ Option (d) also incorrect because y value is > 10/3

∴ Therefore option (a) is correct.

f(x, y) = 3x + 6y = 3 × 4 + 6 × 10
3 3

= 4 + 20 = 24 maximum.

23. The difference between compound interest and simple interest

on an amount of Rs. 15,000 for 2 years is Rs. 96. What is the

rate of interest per annum?

a) 8

b) 10

c) 12

d) Cannot be determined

Answer: (a)

Solution:
�15000 × �1 + 1R00�2 − 15000� – �1500100×0R ×2� = 96

18


⟹ 15000 ��1 + 1R00�2 − 1 − 120R0� = 96

⟹ 15000 �(100 + R)2−1000−(200 × R)� = 96
10000

⟹ R2 = �963× 2� = 64 ⟹ R = 8.

∴ Rate = 8%

24. If |4x – 7| = 5 then the values of 2|X| – |– X| is:

a) 2, 1/3

b) 1/2, 3

c) 3/2, 9

d) 2/3, 9

Answer: (b)

Solution:

4x – 7 = ± 5

4x – 7 = 5

4x = 12

x=3

4x – 7 = – 5

4x = 2 ⇒ x = 1
2

substitute x value in 2|X| – |– X|

(i) x = 1
2

2�21� − �− 12� = 1 – 1 = 1
2 2

(ii) x = 3

2|3| – |– 3| = 6 – 3 = 3

∴ Answer (b) is correct.

19


25. Find the value of
�1 − 13� �1 − 41� �1 − 15� ……………………….. �1 − 1100�
a) 1/40

b) 1/45

c) 50

d) 1/50

Answer: (d)

Solution:

Given expression = 2 × 3 × 4 × … … … … … .× 99
3 4 5 100

= 2 = 1
100 50

26. A firm is selling its product at Rs. 60 per unit. The total cost of

production is Rs. 100 and firm is earning total profit of Rs. 500.

Later, the total cost increased by 30%. By what percentage the

price should be increased to maintained the same profit level.

a) 5

b) 10

c) 15

d) 30

Answer: (a)

Solution:

Let the no. of units = q

Selling price (S.P) = 60/unit

(S.P) total selling price = 60 q

(C.P) total cost price = 100

20


Total profit = 500

Profit = S.P – C.P

= 60q – 100 = 500

60q = 600

∴ No. of units = 10
∴ S.P = 60 × 10 = 600

Total cost increased by

30% = C.P = 130

Total S.P = x

Profit = 500 (same)

Profit = S.P – C.P = 500

x – 130 = 500

x = 630
∴ The difference of

S.P = 630 – 600 = 30
∴ The price should be increased to maintain the same profit

level

30 × 100 = 5%
600

27. �xxqr�(q+r−p) ∙ �xxpr �(r+p+q) ∙ �xxpq�(p+q−r) = ?

a) x(a – b – c)

b) 0.5

c) 1

d) x(a + b + c)

Answer: (c)

21


Solution:
�xxqr�(q+r−p) ∙ �xxpr �(r+q−q) ∙ �xxpq�(p+q−r)
= �x(q−r)�(q+r−p) ∙ �x(r−p)�(r+q−q) ∙ �x(p−q)�(p+q−r)
= x[(q - r)(q + r - p) + (r - p)(r + p - q) + (p - q)(p + q - r)]
= x[(q - r)(q + r)- p(q - r) + (r - p)(r + p) - q(r - p) + (p - q)(p + q) - r(p - q)]
= x[(q2 – r2) – p(q – r) + (r2 – p2) – q(r – p) + (p2 – q2) – r(p – q)]
= x[(q2 – r2 + r2 – p2 + p2 – q2) – p(q – r) – q(r – p) – r(p – q)]
= x[0 - p(q - r) - q(r - p) - r(p - q)]
= x[- p(q - r) - q(r - p) - r(p - q)]
= x[-pq + pr -qr + pq - rp + rq ]
= x[0]
=1
28. A person can row 7(1/2) km an hour in still water. Finds that it
takes twice the time to row upstream than the time to row
downstream. The speed of the stream is:
a) 2 kmph
b) 2.5 kmph
c) 3 kmph
d) 4 kmph
Answer: (b)
Solution:
Let the distance covered be x km and speed of stream = y kmph.
Speed downstream = (15/2) + y = kmph
Speed upstream = (15/2) – y = kmph

22


According to question,
[2x/((15/2) + y)] = [x/((15/2) – y)]
Or, 15 – 2y = (15/2) + y
Or, 3y = 15 – (15/2) = 15/2
Or, y = 15/6 = 2.5 kmph.
29. A train met with an accident 120 km from station A. It
completed the remaining journey at 5/6 of its previous speed and
reached 2 hours late at station B. Had the accident taken place
300 km further, it would have been only 1 hour late. What is the
speed of the train?
a) 100 km/h
b) 12 km/h
c) 60 km/h
d) 50 km/h
Answer: (c)
Solution:
A ____100 km____ P1 ________300 km_____ P2____X______
B
if speed becomes 5/6 then time taken will be 6/5 of original
time.
Thus, extra time = 2 hour
1/5 = 2 hour
Thus, unusual time which is require to complete the journey

= 5 × 2 = 10 hours.

23


It means it covers 300 km distance in 5 hours.
Then speed = 300/5 = 60 kmph.
30. A group of men decided to do a job in 4 days. But since 20
men dropped out every day, the job completed at the end of the
7th day. How many men were there at the beginning?
a) 240
b) 140
c) 280
d) 150
e) None of these
Answer: (b)
Solution:
Let X be the initial number of men then,

According to the question,
4X = X + (X – 20) + (X – 40) + (X – 60) + (X – 80) + (X –
100) + (X – 120)
⟹ 4X = 7X – 420
⟹ 3X = 420
⟹ X = 420/3
⟹ X = 140 men.

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